Question

Two identical metal balls with unequal positive charges are touched. Charge on the first ball decreased by 20 percent. Then the ratio of charge on first ball to second ball before touching was:

• A. $\frac{3}{2}$
• B. $\frac{5}{3}$
• C. $\frac{4}{3}$
• D. $\frac{5}{2}$

Answer 1

Answer: (B)

$\frac{5}{3}$

Answer 2

Let the initial charges on the two identical metal balls be $q_1$ and $q_2$.

When the two identical metal balls are touched, charge is redistributed until the potential on both balls is the same. Since the balls are identical and conducting, the charge will be distributed equally on both balls. The total charge before touching is $Q_{total} = q_1 + q_2$. After touching, the charge on each ball becomes equal to the total charge divided by 2, so $q'_1 = q'_2 = \frac{q_1 + q_2}{2}$.

The problem states that the charge on the first ball decreased by 20 percent. This means the final charge on the first ball ($q'_1$) is 20 percent less than the initial charge on the first ball ($q_1$). Thus, $q'_1 = q_1 - 0.20 q_1 = 0.80 q_1$.

Now, we equate the two expressions for the final charge on the first ball: $\frac{q_1 + q_2}{2} = 0.80 q_1$.

Multiplying both sides by 2: $q_1 + q_2 = 1.60 q_1$.

Isolating the terms involving $q_1$ and $q_2$: $q_2 = 1.60 q_1 - q_1$ $q_2 = 0.60 q_1$.

To find the ratio $\frac{q_1}{q_2}$, divide both sides of the equation $q_2 = 0.60 q_1$ by $q_2$: $1 = 0.60 \frac{q_1}{q_2}$.

Solving for the ratio $\frac{q_1}{q_2}$: $\frac{q_1}{q_2} = \frac{1}{0.60} = \frac{1}{\frac{3}{5}} = \frac{5}{3}$.

Therefore, the ratio of charge on the first ball to the second ball before touching was $\frac{5}{3}$.